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SashulF [63]
3 years ago
9

What is 35% of $120?

Mathematics
1 answer:
allsm [11]3 years ago
4 0

Answer:

Step-by-step explanation:

 \frac{35}{100} *120\\=7*6\\=42

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the art club is selling tubo of cookies dough for a fundraiser. how much Will a student earn if he or she sells 24 tus of cookie
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Answer:

if each tub is $12 the student would make $288

Step-by-step explanation:

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At a production process, the produced items are tested for defects. A defective unit is classified as such with probability 0.9,
Snezhnost [94]

Answer:

We use Baye's theorem:  P(A)P(B|A) = P(B)P(A|B)

with (A) being defective and

(B) marked as defective

we have to find P(B) = P(A).P(B|A) + P(¬A)P(B|¬A). .......eq(2)

Since  P(A) = 0.1 and P(B|A)=0.9,

P(¬A) = 1 - P(A) = 1 - 0.1 = 0.9

and

P(B|A¬) = 1 - P(¬B|¬A) = 1 - 0.85 = 0.15

put these values in eq(2)

P(B) = (0.1 × 0.9) + (0.9 × 0.15)

       = 0.225 put this in eq(1) and solve for P(B)

P(B) = 0.4

6 0
2 years ago
Please answer asap. there are two pics :)
Thepotemich [5.8K]

Answer:

\boxed{\sf A. \ 0.34}

Step-by-step explanation:

The first triangle is a right triangle and it has one acute angle of 70 degrees.

We can approximate \sf \frac{WY}{WX} from right triangle 1.

The side adjacent to 70 degrees is WY. The side or hypotenuse is WX.

The side adjacent to 70 degrees in right triangle 1 is 3.4. The side or hypotenuse is 10.

\sf \frac{3.4}{10} =0.34

8 0
3 years ago
In a certain population of the eastern thwump bird, the wingspan of the individual birds follows an approximately normal distrib
Greeley [361]

Answer:

a) P(48 < x < 58) = 0.576

b) P(X ≥ 1) = 0.9863

c) E(X) 2.88

d) P(x < 48) = 0.212

e) P(X > 2) = 0.06755

Step-by-step explanation:

The mean of the wingspan of the birds = μ = 53.0 mm

The standard deviation = σ = 6.25 mm

a) Probability of a bird having a wingspan between 48 mm and 58 mm can be found by modelling the problem as a normal distribution problem.

To solve this, we first normalize/standardize the two wingspans concerned.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ

For wingspan 48 mm

z = (48 - 53)/6.25 = - 0.80

For wingspan 58 mm

z = (58 - 53)/6.25 = 0.80

To determine the probability that the wingspan of the first bird chosen is between 48 and 58 mm long. P(48 < x < 58) = P(-0.80 < z < 0.80)

We'll use data from the normal probability table for these probabilities

P(48 < x < 58) = P(-0.80 < z < 0.80) = P(z < 0.8) - P(z < -0.8) = 0.788 - 0.212 = 0.576

b) The probability that at least one of the five birds has a wingspan between 48 and 58 mm = 1 - (Probability that none of the five birds has a wingspan between 48 and 58 mm)

P(X ≥ 1) = 1 - P(X=0)

Probability that none of the five birds have a wingspan between 48 and 58 mm is a binomial distribution problem.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan between 48 mm and 58 mm = 0

p = probability of success = Probability of one bird having wingspan between 48 mm and 58 mm = 0.576

q = probability of failure = Probability of one bird not having wingspan between 48 mm and 58 mm = 1 - 0.576 = 0.424

P(X=0) = ⁵C₀ (0.576)⁰ (1 - 0.576)⁵ = (1) (1) (0.424)⁵ = 0.0137

The probability that at least one of the five birds has a wingspan between 48 and 58 mm = P(X≥1) = 1 - P(X=0) = 1 - 0.0137 = 0.9863

c) The expected number of birds in this sample whose wingspan is between 48 and 58 mm.

Expected value is a sum of each variable and its probability,

E(X) = mean = np = 5×0.576 = 2.88

d) The probability that the wingspan of a randomly chosen bird is less than 48 mm long

Using the normal distribution tables again

P(x < 48) = P(z < -0.8) = 1 - P(z ≥ -0.8) = 1 - P(z ≤ 0.8) = 1 - 0.788 = 0.212

e) The probability that more than two of the five birds have wingspans less than 48 mm long = P(X > 2) = P(X=3) + P(X=4) + P(X=5)

This is also a binomial distribution problem,

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan less than 48 mm = more than 2 i.e. 3,4 and 5.

p = probability of success = Probability of one bird having wingspan less than 48 mm = 0.212

q = probability of failure = Probability of one bird not having wingspan less than 48 mm = 1 - p = 0.788

P(X > 2) = P(X=3) + P(X=4) + P(X=5)

P(X > 2) = 0.05916433913 + 0.00795865476 + 0.00042823218

P(X > 2) = 0.06755122607 = 0.06755

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